Area of a circle which is interior to the parabola

Question

I'm trying to solve the following problem:

Find the area of the circle $x^2+y^2=8$, which is interior to the parabola $y^2=2x$.

I have my own solution and I want to verify whether it is correct. My solution is to take the $1/4$ of the area of the circle, which is $\int_{0}^{\sqrt{8}}\sqrt{-x^2+8}$ and subtract the rest of the area we don't want, which is : $\int_{0}^{2}\sqrt{-x^2+8} -\sqrt{2x}$ (it is the 1/4 of the area of the circle minus the area under the parabole).

Therefore the area is $2*(\int_{0}^{\sqrt{8}}(\sqrt{-x^2+8} )- \int_{0}^{2}(\sqrt{-x^2+8} -\sqrt{2x}))$.

Is my solution correct?

Thanks

Answer 1

Take a look at the picture below:

The area that you're after is the area of a quarter of the given circle (in this case, that's $2\pi$) plus twice the area of the region below the parabola and above the line $y=x$.

So, the area is equal to$$2\pi+2\int_0^2\sqrt{2x}-x\,\mathrm dx=2\pi+\frac43.$$

Answer 2

Looking at it graphically we can notice that this area is equal to the area of a semicircle minus the area under the quadratic: $$A=\frac{\pi}{2}r^2-\int_{-r}^rxdy=\frac{\pi}{2}r^2-\int_{-r}^r\frac{y^2}{2}dy$$ Now just work this out and sub in $r=2\sqrt{2}$